command concatenation in bash - bash

Question

I want to invoke a function with some input checks (the input should be an integer between 1-21). If ok, then do the echo "invoke", else just print a message about invalid input.
I tried with the following simplified example, it works for the invalid case, but does not invoke for the valid case. what is wrong?
function _check_num ()
{
[[ "$1" =~ ^[0-9]+$ ]] && [ "$1" -ge 1 -a "$1" -le 21 ] || echo "input should be (1-21)" && return 1 // one-liner
}
function _call()
{
_check_num $1 && echo "invoke only if input is 1-21" // does not invoke given valid input
}
Note: please explain me the root cause of this one-liner case.

Answer 1

Do not use chains of && and || as a replacement for an if statement.
&& and || have equal precedence, so a && b || c runs c if either a or b fail; it is not equivalent to if a; then b; else c; fi.
a && b || c && d is parsed as ((a && b) ||c) && d, not (a && b) || (c && d).
Use an expicit if statement to make your code readable. (Also, don't use -a inside [...]; it is considered ambiguous and obsolete.)
function _check_num ()
{
if [[ "$1" =~ ^[0-9]+$ ]] && [ "$1" -ge 1 ] && [ "$1" -le 21 ]; then
return 0
else
echo "input should be (1-21)" >&2
return 1
fi
}
The less readable version would be something like the following, uses braces to properly group the commands.
function _check_num ()
{
{
[[ "$1" =~ ^[0-9]+$ ]] && [ "$1" -ge 1 ] && [ "$1" -le 21 ]
} || {
echo "input should be (1-21)" >&2 && return 1
}
}

Answer 2

function _check_num ()
{
if [[ "$1" =~ ^[0-9]+$ ]] && [ "$1" -ge 1 -a "$1" -le 21 ]; then
return 0
else
echo "input should be (1-21)"
return 1
fi
}
I thin you missed your return 0 statement
See chepner answer: problem is priority of && / || operators
Normally the return status of the first part should be 0 but idk something could go wrong.

Answer 3

You can use group last echo and return in {...} to make it work:
function _check_num () {
[[ "$1" =~ ^[0-9]+$ ]] && [ "$1" -ge 1 -a "$1" -le 21 ] ||
{ echo "input should be (1-21)" && return 1; }
}
Without {...} last return 1 is always returning 1 whether value is valid or invalid.

Answer 4

If you put your "one liner" inside a script to test it:
function _check_num () {
[[ "$1" =~ ^[0-9]+$ ]] &&
[ "$1" -ge 1 -a "$1" -le 21 ] ||
echo "input should be (1-21)" &&
return 1 # one-liner
}
_check_num "$1"; echo "exit value $?"
we get:
$ ./script.sh qwe
input should be (1-21)
exit value = 1
$ ./script.sh 112
input should be (1-21)
exit value = 1
$ ./script.sh 12
exit value = 1
As you can see, the exit value is always 1 (not a successful result).
Therefore you can not use the exit value as the trigger for other code.
Where?
The core issue is in this structure:
[ "$1" -ge 1 -a "$1" -le 21 ] || echo "input should be (1-21)" && return 1
which could be reduced to:
[ … ] || echo && return 1
The two possible exit values (success or not) of the […] could be tested with:
f(){ true || echo && return 1; }; f; echo "$?" # prints a 1.
f(){ false || echo && return 1; }; f; echo "$?" # *also* a 1[1].
[1] After also printing a blank line from the internal echo.
Why ?
Because of the "short circuit" effect of the shell «AND and OR lists»
From bash manual:
Lists
Of these list operators, && and || have equal precedence …
The return status is the exit status of the last command executed.
AND and OR lists are sequences of one of more pipelines separated by the && and || control operators, respectively.
What is crucial to understand the issue is this part (for AND):
An AND list has the form
command1 && command2
command2 is executed if, and only if, command1 returns an exit status of zero.
And for OR
An OR list has the form
command1 || command2
command2 is executed if and only if command1 returns a non-zero exit status.
And the natural understanding of which should be the present exit status at any position:
The return status of AND and OR lists is the exit status of the last command executed in the list.
If you split any AND and OR list at any position, the exit status at such position is the exit status of the last previous command executed.
Some commands may get bypassed.
Step by step:
The first command is a test [[…]], the exit value at this point is its exit value.
This first command is connected to the next with an OR (||).
If the exit value of the first command is fail (not 0) the next command (echo) will be executed.
The exit code of echo is always success.
The next connection is an AND (&&).
At that point the exit value is true, the next command will be executed.
If the exit value of the first command is success (0) the next command will not be executed.
But the next one (return 1) will because the connection is an AND.
In both cases the return 1 is the last command executed.
The return value is always 1.
Precedence
Precedence may affect the order of the commands executed, but that does not explain why some commands are not executed.
Precedence in an "AND and OR list" is the same for && and ||.
So, operators will be considered in the left to right order they are found.
Associativity
Associativity in a shell "AND and OR list" is "left-associative".
That means that: if there are no parenthesis, operations are grouped from the left.
The first command is grouped with the second.
The result of that is grouped with the third, etc.
But even grouped as explained here. That does not explain why some commands are bypassed.
References
Advanced Bash-Scripting Guide: Chapter 26. List Constructs
Precedence
Operator associativity
Maybe you should consider using a function like:
function _check_num () {
declare -i num
if [[ $1 =~ ^[+]?(0+)?([0-9]+)$ ]]; then
num=${BASH_REMATCH[2]}
if ! (( num >=1 && num <= 21 )); then
echo "input should be (1-21)"
return 1
fi
return 0
else
echo "input should be a number"
return 2
fi
}